(6 pts) In the circuit below, the switch has been closed for a long time and is then opened at t = 0. Pay attention to the polarity of the capacitor. a) What is the charge on the capacitor just before the switch is closed? b) What is the charge on the capacitor after the switch has been closed a long time? c) By how much does the charge on the capacitor change?
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Given: Capacitance, C = 10 µF Potential difference, V = 3 V Using the formula Q = C * V, where Q is the charge on the capacitor: Q = 10 * 10^-6 * 3 Q = 30 * 10^-6 Q = 30 µC Show more…
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